Question

Question 2
Given that the zeroes of the cubic polynomial $x^3–6x^2+3x+10$ are of the form (a), (a + b) and (a + 2b) for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

Answer 1

Let $f\left(x\right)=x^3–6x^2+3x+10$
Given that, a, (a + b) and (a + 2b) are the zeroes of f(x) Then,
sum of zeros = $\frac{-b}{a}$
$\Rightarrow a+(a+b)+(a+2b)=-\frac{(-6)}{1}$
$\Rightarrow 3a+3b=6$
$\Rightarrow a+b+2$
Sum of product of two zeroes at a time =$\frac{c}{a}=\left(\frac{coefficientofx}{coefficientof{x}^3}\right)$
$\Rightarrow a(a+b)+(a+b)(a+b)+a(a+2b)=\frac{3}{1}$
$\Rightarrow a(a+b)+(a+b)(a+b)+b+a(a+b)+b=3$
$\Rightarrow 2a+2(2+b)+a(2+b)=3$ [ using Eq. (i)]
$\Rightarrow 2a+2(2+2–a)+a(2+2–a)=3$ [using Eq. (i)]
$\Rightarrow 2a+8–2a+4a–a^2=3$
$\Rightarrow -a^2+8=3–4a$
$\Rightarrow a^2–4a–5=0$
Using factorization method
$a^2–5a+a–5=0$
$\Rightarrow a(a–5)+1(a–5)=0$
$\Rightarrow (a-5)(a+1)=0$
$\Rightarrow a=-1,5$
When a = - 1 , then b = 3
When a = 5, then b = - 3
$\therefore$ Required zeroes of f(x) are
When a = - 1 and b = 3
Then, a, (a + b), (a + 2) = - 1, ( - 1 + 3), ( - 1 + 6) or – 1, 2, 5
When a = 5 and b = - 3, then
a, (a+ b), (a + 2b) =5, (5 – 3), (5 – 6) OR (5, 2, -1).
Hence, the required values of a and b are a = -1 and b = 3 or a = 5, b = -3,
$andthezeroesinbothcasesare-1,2and5$.